Optics Question 869

Question: Two identical coherent sources are placed on a diameter of a circle of radius R at separation x (« R) symmetrical about the center of the circle. The sources emit identical wavelength $ \lambda $ each. The number of points on the circle of maximum intersity is $ (x=5\lambda ) $

Options:

A) 20

B) 22

C) 24

D) 26

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Path difference at P is $ \Delta x=2( \frac{x}{2}\cos \theta )=x\cos \theta $

For intensity to be maximum, $ \Delta x=n\lambda $ (n=0, 1, 2, 3, ….) or $ x\cos \theta =n\lambda $

or $ x\cos \theta =\frac{n\lambda }{x}\ge 1 $

$ \therefore n\ge \frac{x}{\lambda } $

Subsituting x= 51, we get $ n\ge 5 $ or n= 1, 2, 3, 4, 5…….

Therefore in all four quadrants there can be 20 maxima.

There are more maxima at $ \theta =0{}^\circ $

and $ \theta =180{}^\circ $ .

But n = 5 corresponds to $ \theta =90{}^\circ $

and $ \theta =270{}^\circ $ which are coming only twice while we have multuplies it four times.

Therefore, total number of maxima are still 20, i.e., n = 1 to 4 in four quadrants (total 16) plus more at $ \theta =0{}^\circ $ , $ 90{}^\circ $ , $ 180{}^\circ $ and $ 270{}^\circ $ .



NCERT Chapter Video Solution

Dual Pane